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The determination of a rate equation

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The determination of a rate equation Introduction After watching the demonstration shown by the teacher, I noticed a cloudy solution forming within 10 seconds after mixing the 50cm� of (0.4 moldm-�) Na2S2O3 was mixed with 5.0 cm� HCL (2.0 moldm-�) diluted with 20cm� H2O. From this I can simply conclude that the reaction takes place quickly and further increasing concentration/ volumes would produce a faster reaction. This would make it difficult to measure the exact point when a cloudy solution is formed. If the time taken to produce the cloudy solution is faster than 10seconds this increases the chance of error and inaccuracy as the time taken for detection from the eye and time to respond by stopping the stop clock can lead to an error of judgement within +5 seconds, this would be too great if the reaction was carried out at a greater concentration/ volumes. I know SO2 is produced from the equation below which is a gas: 2HCL + Na2S2O3 2NaCl + SO2 + S + H2O After a while I noticed a smell of SO2 as it diffused across the room. This is important, if using large amounts of Na2S2O3 this will cause the production of more SO2 which can be dangerous and difficult to get rid of. If I decide to use 5cm� of Na2S2O3 and 1cm� of HCL the reactions would be slow and varying the concentration would result in difficulty and a greater chance of error as measuring decimal places for reactants using equipment has greater percentage error Therefore after consideration and thought of all these factors I have decided to keep a total volume of 20cm� this will enable me to vary the concentration of HCL and Na2S2O3 and obtain suitable results via dilution using water. I will use 5 different concentration of one reactant and keep the other reactant constant. In using 5 concentrations this will give me enough pieces of data to plot a safe graph. ...read more.


E.g. 1st test tube, using the burette to measure 2 cm� of HCl, then using the 10cm� measuring cylinder add required amount of water (8 cm�). Label the test tube. 10. Now repeat this procedure and vary the amount of HCl as decided (4, 6, 8, 10 cm�) into separate test tubes and add correct amounts of water (6, 4, 2, 0 cm�) and label each test tube with a number. 11. On the 1st test Tube stick a small piece of paper containing a cross (X) marked on it using a ball point pen. (on the side of the test tube as shown in the diagram, so it is covered by the solution) 12. For the first experiment you will need to measure out 10 cm� of Na2S2O3 from the second burette into a clean test tube, as this will be a constant when determining the rate for HCl. 13. Add the Na2S2O3 to the HCl in the 1st test tube and at the same time start the stop watch. 14. An end point will be determined when sulphur precipitate is formed. This will turn the solution from clear colourless to cloud and the cross (X) marked on the test tube will not be visible. When this point is reached you must immediately stop the watch and record the time in a table as shown below: Test tube No: Concentration of HCL (moldm-�) Time (seconds) 1 0.4 2 0.8 3 1.2 4 1.6 5 2.0 15. Repeat the experiment again but this time add the 10 cm� Na2S2O3 into the 2nd test tube and record the time. Continue each repetition at the different concentrations and complete the table above. 16. After obtaining all the readings for HCl, rinse out the test tubes and prepare them using Na2S2O3 and label them as before. (2, 4, 6, 8, 10 cm� of Na2S2O3 into each test tube). ...read more.


As I carried out the experiment in test tubes each time, this kept it fair as I looked through the same amount of solution each time. However I noticed especially when adding the Na2S2O3 to the HCl that the solutions did not mix properly, as I observed a solid precipitate forming 3/4th down the test tube. This indicated that a sulphur precipitate formed and the whole of the solutions did not react. Therefore I repeated the experiment but this time I gently moved the test tube to help mix the reactants. This error in shaking test tubes could have led to some difference in results as I did not shake all the test tubes. As I know if I shake the test tube I am providing more kinetic energy, more molecules collide due to energy provided to molecules. This could be overcome by using glassware that has a greater surface area, such as 50cm� measuring cylinder or a conical flask. In the method there is no indication of replication, but I have repeated the experiment twice due to the time available and the amount of reactants. (I have only accounted for 1 replication in the method concerning the quantity of each solution, therefore multiply each solution quantity by the number of replications carried out in the future). I would like to have repeated the experiment once more to achieve more accurate and reliable results. Finally, form the results obtained I can say that I have no anomalous results, the graphs produced are sufficient enough to produce the rate equation for this reaction. If I repeated the experiment once more and at more concentrations (e.g. HCl 0.2, 0.4, 0.6, 0.8, 1.0, 1.2, 1.4, 1.6, 1.8, 2.0moldm-�) this would give me a greater confidence in my results and their reliability, as the graph produced would contain more data that would provide a detailed accurate account of the rate. ?? ?? ?? ?? Abdul Basit Farhat Chemistry 1 ...read more.

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